Question

Problem 1. (point) A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring feet. The ball is started in motion from the equilibrium position with a downward velocity of 9 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity on feet per second). Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that this means that the positive direction for y is down) ye Take as the gravitational acceleration 32 feet per second per second. Problem 2. (1 point) If the differential equation M + 6 + 9x0 d! is overdamped, the range of values for mis? Your answer will be an interval of numbers given in the form (1,2), (1.2).(-int, 6), etc. Pr Tent

Answer 1

Given mass of the ball, m = 128 pounds

Yo = 128/485 feet

Air resistance, Fv = 4dy/dt

We need to find the position of ball, y below the rest position in terms of t.

The solution is shown in the figure:

Problem 1 weight of the ball = 128 pounds, Equilibrim disp., yo ft. s 128 485 Initial donnard velocity, Vi = 9 ft.1sec Avi resistance in pounds) 4 dy be the position of the ball belone equilibruin as positive Atequilibrum, mg =kky 128x 32= KX 128 Tet, y 485 >> k 485 x 32 IbTft

/ Thus, spring constant, K = 425*32 1618 Using Neneton's 2nd Laws of motion : may = ky+ = ky+ zdy at2 dt +KA=0 st --> more ada de but, m=12816 K: 485*32 1b/ft => 1128 dy + 204 #485532) 420 Nors, ch'Brocteristics equationes: (1280? +20 + 485*32)X =0 => 122 m2 + 2m + 445*32 >m: -0.0078 I1100114 Hore, the roots are imaginary y = ce (-240078t; (, cos (11.0114 t) +(zsen (11.01141)) sere, y (0) = 128 = e (c) => a G = 0.264 And, y los = = 9 ~ y' (t) = +0.0070) @ 0001? t (sos (1101146)+cz sencil.01144)) (-11.0114 C, Sun (17.0114+)+11:01110 ( -0.0022+) 123 485 485 Totoong COS(110 Y60) -9 = (-0.0078)(,) + (-0.0078)((,) + ( + 1l. 0114 (2) - (-0.0076X0.264) + 11.0114 Cz 0.8175] و > 2 1-0*6018+) (O 01264 cos (11.01146) +0.8175 Sun (11.01144) Sc, 4(+)- e

6 dx make 9x=0 2 & de dret + c Problem 2 Cuven 6 dr m du dt 9 dt² This is analogous to- o 2 dt² n at And equation ( is also analogousto- m det a da k dt² wher, mem, ko9, 236, Thus, damping wef. Ee = a 2 km For overdamped vibrations, Ze >O K at 2017 2om >o > zom Gm

So, the range of values of mis is : ECO, mis always postie

So, the range of values of mis is : ECO, mis always postie